Proving that $\mathbb{Z_{-}} \cap \mathbb{N}=\emptyset$

Question

In assumption that $\mathbb{N}$ and successor function ($\overline{x}$) over $\mathbb{N}$ is defined by 5 Peano axioms:

1. $1\in\mathbb{N}$
2. $n\in\mathbb{N} \Rightarrow \overline{n}\in\mathbb{N}$
3. $\nexists n\in\mathbb{N}:\ \overline{n}=1$
4. $\forall n,m,k\in\mathbb{N}:\ (\overline{n}=k\land\overline{m}=k)\ \Rightarrow\ n=m$
5. $\left[P(1)\land\forall n\in\mathbb{N}\ [P(n)\Rightarrow P(\overline n)]\right]\ \Rightarrow \ \forall n\in\mathbb{N}\ [P(n)]$

and an addition operation is defined over $\mathbb{N}$ by:

1. $\forall n\in\mathbb{N}\quad n + 1 = 1 + n = \overline{n}$
2. $\forall n,m\in\mathbb{N}\quad n+\overline{m}=\overline{n+m}$

and that we have proved some base properties of addition over $\mathbb{N}$ like existence, uniqueness and commutivity, we are extending $\mathbb{N}$ and operations to new set (called $\mathbb{Z}$) by including a neutral element (denoted by "$0$") where: $$\tag{1}\label{d:zero}\forall n\in \mathbb{N}\quad n+0=0+n=n$$ *and we know that $0\notin\mathbb{N}$, because if $0\in\mathbb{N}$ then by substitution $n=1$ to \eqref{d:zero}: $$0+1=1\quad\Leftrightarrow\quad \overline{0}=1\quad\text{(and that one contradicts 3rd Peano axiom)}$$ and by constructing negative numbers $(-n)\in\mathbb{Z}_-$ for all $n\in\mathbb{N}$ by relation $n+(-n)=0$.

So the question is how to prove that constructed set $\mathbb{Z}_-$ is not already in $\mathbb{N}$? Maybe there is a simple prove for $(-1)$ which I can extend to other numbers.

Answer 1

Suppose $-x \in \mathbb{N}$ for some $x \in \mathbb{N}$. Then $0 = x + (-x) \in \mathbb{N}$ (because $\mathbb{N}$ is closed under addition). But as you already showed, $0 \notin \mathbb{N}$. Contradiction!